thiêt ke he t-d cho may bào giường

7/27/2019 thit ke he T-D cho may bo ging

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MC LC

CHNG I. TNG QUAN V CNG NGH MY BO GING V LA CHNPHNG N THAY TH ................................................................................................. 4

1. Tng quan v my bo ging .................................................................................... 4

1.1. Gii thiu v cng ngh ........................................................................................ 41.2. Cc yu cu i vi h thng truyn ng in my bo ging ....................... 7

2. La chn phng n thay th h truyn ng cho my bo ging 7210 .................. 9

3. Chn phng n mch lc ......................................................................................... 10

3.1. Chnh lu hnh tia ba pha ................................................................................... 103.2. Chnh lu cu mt pha ....................................................................................... 113.3. Chnh lu cu ba pha ......................................................................................... 12

CHNG II. THIT K MCH LC ............................................................................ 14

1. Nguyn l mch lc .................................................................................................... 14

1.1. S ................................................................................................................... 141.2. Gin in p, dng in ................................................................................ 141.3. Nguyn l lm vic v cc biu thc tnh ton ................................................... 151.4. S mch lc ................................................................................................... 16

2. Thit k bin p lc ................................................................................................... 17

2.1. Tnh cng sut my bin p lc .......................................................................... 172.2. Thit k bin p .................................................................................................. 18

3. Chn van thyristor ..................................................................................................... 24

3.1. Chn van theo ch tiu dng in ....................................................................... 243.2. Chn van theo ch tiu in p ........................................................................... 25

4. Tnh chn cun khng san bng ................................................................................ 26

4.1. Tnh gi tr cun khng lc ................................................................................ 264.2. Thit k cun khng lc ...................................................................................... 29

5. Tnh ton bo v mch lc ........................................................................................ 32

5.1. Bo v qu dng ................................................................................................. 32


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5.2. Bo v qu in p ............................................................................................. 34CHNG III. THIT K MCH TO XUNG IU KHIN ...................................... 36

1. Khu ng b ............................................................................................................. 37

1.1. S (hnh 3.2a) ................................................................................................ 371.2. Gin xung (hnh 3.2b) .................................................................................... 371.3. Gii thiu s .................................................................................................. 371.4. Nguyn l lm vic .............................................................................................. 37

2. Khu to in p ta .................................................................................................. 38


3. Khu so snh .............................................................................................................. 40


4. Khu trn xung v khuch i xung (KX) ............................................................. 41

4.1. Khu trn xung ................................................................................................... 414.3. S ca khu trn xung v khuch i xung ................................................... 44

5. Tnh ton thng s mch iu khin .......................................................................... 45

5.1. Tnh ton khu khuch i xung ......................................................................... 465.2. Tnh ton khu trn xung ................................................................................... 475.3. Tnh ton khu so snh ....................................................................................... 485.4. Tnh ton khu to in p ta ........................................................................... 486.5. Tnh ton khu to xung ng b ....................................................................... 49

CHNG IV. TNG HP H V M PHNG .......................................................... 50

1. Tnh ton thng s ..................................................................................................... 50


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1.1. ng c .............................................................................................................. 501.2. B bin i .......................................................................................................... 51

2. Tng hp mch vng dng in ................................................................................ 51

3. Tng hp mch vng tc ...................................................................................... 54

4. M phng h thng .................................................................................................... 57

TI LIU THAM KHO ................................................................................................. 61


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CHNG I. TNG QUAN V CNG NGH MY BO GING V

LA CHN PHNG N THAY TH

1. Tng quan v my bo ging

1.1. Gii thiu v cng nghMy bo ging l loi my c th gia cng cc chi tit ln, chiu di bn c th t

1,5 m n 12 m. Tu thuc vo chiu di ca bn my v lc ko c th phn loi my

thnh ba loi :

+My c nh : chiu di bn Lb < 3 m, lc ko Fk= 30 50 kN.

+My c trung bnh : chiu di bn Lb = 4 5 m, lc ko Fk= 50 70 kN.

+My c nng : chiu di bn Lb > 5 m, lc ko Fk> 70 kN.

Hnh 1.1. Dng bn ngoi ca my bo ging.

Trong :

1: chi tit gia cng

2: bn my

3: dao ct

4: bn dao ng

5: x ngang c nh


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Chi tit gia cng c kp cht trn bn my v c th chuyn ng tnh tin qua

li. Dao ct c kp cht trn bn dao ng, bn dao ny c t trn x ngang c nh

khi gia cng. Trong qu trnh lm vic bn my di chuyn qua li theo cc chu k lp li,

mi chu k gm hai hnh trnh thun v ngc.-Hnh trnh thun thc hin gia cng chi tit gi l hnh trnh ct gt.

-Hnh trnh ngc bn my chy v v tr ban u, khng ct gt gi l hnh trnh

khng ti.

C sau khi kt thc hnh trnh ngc bn dao li di chuyn theo chiu ngang mt

khong gi l lng n dao s ( mm/ hnh trnh kp ).

Chuyn ng qua li ca bn my gi l chuyn ng chnh.

Dch chuyn ca bn dao sau mi hnh trnh kp gi l chuyn ng n dao.

Chuyn ng ph l chuyn ng nhanh ca x, bn dao, nng u dao trong hnh

trnh khng ti.

Hnh 1.2. th tc ca bn my

Gi thit bn my ang u hnh trnh thun v c tng tc n tc

v0=515 m/pht trong khong thi gian t1.


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- t1: khong thi gian tng tc n v0 = 515 m/pht.

- t2: khong thi gian chy n nh vi tc v0, sau t2 dao ct vo chi tit

-t3: bn my tip tc chy vi tc n nh v0

-t4: tng tc t v0vth

-t5: bn my chuyn ng vi tc Vth v thc hin gia cng chi tit

-t6: bn my s b gim tc n V0

-t7: bn my lm vic n nh vi tc ca bn my l v0

-t8: dao c ra khi chi tit khi tc ca bn my l v 0

-t9,t10: o chiu t hnh trnh thun sang hnh trnh ngc n tc v ng

-t11: bn my chy theo hnh trnh ngc vi tc vng

-t12: thi gian gim tc n v0 hnh trnh ngc

-t13: bn my chy n nh tc thp v0 chun b o chiu

-t14: o chiu sang hnh trnh thun bt u thc hin 1 chu k khc.

Bn dao c di chuyn bt u t thi im bn my o chiu t hnh trnh

ngc sang hnh trnh thun v kt thc di chuyn trc khi dao ct vo chi tit.

Tc hnh trnh thun Vth c xc nh tng ng bi ch ct,thng V2 = 5

n (75120) m/ph, tc gia cng ln nht c th t (75120) m/ph. tng nng sut

ca my tc hnh trnh ngc thng c chn ln hn tc hnh trnh thun

Vng=k.Vth (thng k=23).

Nng sut ca my ph thuc vo s hnh trnh kp trong mt n v thi gian:

ngthck ttTn

11

Trong :

Tckthi gian ca mt chu k lm vic ca bn my [s]

tth thi gian bn my chuyn ng hnh trnh thun [s]

tng - thi gian bn my chuyn ng hnh trnh ngc [s]


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Gi L l chiu di hnh trnh bn my:

dcng

dcngth tV

LktVLVLn

).1(1

//

1

Trong :

th

ng

V

Vk -t s gia tc hnh trnh thun v hnh trnh ngc.

tc : thi gian o chiu ca my

Nng sut ca my ph thuc vo k v tc:

+ Khi tng k th nng sut ca my tng, nhng khi k > 3th nng sut tng khng

ng k do tc tng.+ Khi Lb > 3 th tc t nh hng, nng sut ph thuc ch yu vo k.

+ Khi Lbb vth = (75 120) m/ph th tc nh hng nhiu ti nng sut.

V vy, khi thit k truyn ng chnh my bo ging cn phn u gim thi gian

qu trnh qu .

1.2. Cc yu cu i vi h thng truyn ng in my bo ging

1.2.1. Truyn ng chnhPhm vi iu chnh tc truyn ng chnh l t s gia tc ln nht ca bn

my (tc ln nht trong hnh trnh ngc) v tc nh nht ca bn my (tc

thp nht trong hnh trnh thun).

D =min

max

V

V=

min.

max.

th

ng

V

V

Trong :

Vng.max tc ln nht ca bn my hnh trnh ngc, thng

Vng.max = 75 120 m/pht.

Vth.min tc nh nht ca bn my trong hnh trnh thun,

thng Vth.min = 4 6 m/pht


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Nh vy D = (12,5 30) / 1.

Hnh 1-3: th ph ti ca truyn ng chnh MBG

Thng thng, h thng truyn ng in s dng ng c in mt chiu c

cp ngun t b bin i (BB). Theo yu cu ca ph ti tc c iu chnh theo

hai vng:

- Thay i in p phn ng trong phm vi (5 6)/1 vi mmen trn trc ng c

l hng s ng vi tc bn thay i t Vmin=(4 6) m/ph n Vgh = (20 25)

m/ph, khi lc ko khng i.

- Gim t thng ng c trong phm vi (4 5 )/1, khi thay i tc t V gh n

Vmax =(75 120) m/ph, khi cng sut ko gn nh khng i.

Nu s dng phng php iu chnh t thng th s lm gim nng sut my. V

th ngi ta thng m rng phm vi iu chnh in p, gim phm vi iu chnh t

thng, hoc iu chnh tc ng c trong c di bng thay i in p phn ng.

Trong trng hp ny cng sut ca ng c phi tng ln Vmax/Vmin ln.

ch xc lp, n nh tc khng ln hn 5% khi ph ti thay i t 0 n

nh mc. i vi my bo ging c nh h thng truyn ng chnh thng l ng ckhng ng b khp li hp; ng c khng ng b rto dy qun hoc ng c mt

chiu kch t c lp v hp tc . Nhng my c trung bnh th h thng truyn ng l

F-. i vi my c nng h truyn ng l h F- c b khuch i trung gian ; h

chnh lu dng tiristor - ng c mt chiu T-.


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1.2.2. Truyn ng n dao

Truyn ng n dao c tnh cht chu k. Phm vi iu chnh lng n dao l

D=(100 200)/1. Lng n dao cc i c th t ti 100 mm/hnh trnh kp. C cu n

dao lm vic vi tn s ln, c th t ti 1000 ln/gi. H thng di chuyn u dao cnphi m bo theo hai chiu c ch di chuyn lm vic v di chuyn nhanh. Truyn

ng n dao c thc hin bi ng c khng ng b rto lng sc v hp tc .

1.2.3. Truyn ng ph

Truyn ng ph m bo cc di chuyn nhanh bn dao, x my, nng u dao

trong hnh trnh ngc, c thc hin bi ng c khng ng b v nam chm in.

c im truyn ng chnh ca my bo ging l o chiu vi tn s ln,

mmen khi ng, hm m, qu trnh qu chim t l ng k trong chu k lm vic,

chiu di hnh trnh cng gim th nh hng ca qu trnh qu cng tng. V vy,

mun qu trnh qu , khi ng, hm m ta cn chn h thng truyn ng ph hp.

2. La chn phng n thay th h truyn ng cho my bo ging 7210

Truyn ng chnh ca my bo ging s dng h truyn ng F- c tn ti

nhiu nhc im nh :

- S dng nhiu my in quay nn hiu sut thp, cng knh, tn din tch lpt, gy n ln.

- Cng sut t ln, vn u t cao.

- My pht mt chiu c t d nn c tnh t ho c tr kh iu chnh su tc

.

V vy, Thay th h F-D bng h T- h chnh lu iu khin- ng c . B

bin i chnh lu bn dn, bin i trc tip nng lng in xoay chiu thnh nng

lng in mt chiu khng qua khu trung gian no nn c nhiu u im nh:

- tc ng nhanh, khng gy n, d t ng ho do cc van ban dn c h s

khuch i cng sut cao.

- Thun li cho vic thit lp cc h thng t ng iu chnh nhiu vng nng

cao cht lng h thng.


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- Dng cho h c cng sut ln, vn u t v sa cha t hn h F-.3. Chn phng n mch lc

3.1. Chnh lu hnh tia ba pha

Hnh 1.4. S v th lm vic ca chnh lu hnh tia ba pha

Xung iu khin cc van lch pha nhau 1200 in. Gi tr Ud0 = 1,17U2. Vi ti l ng c, in cm Ld ln coi dng id l lin tc v phng, c

quy lut iu chnh:

Ud = Ud0cos

Trong Ud0: tr s trung bnh ca in p chnh lu it.

U2: tr s hiu dng ca in p pha cun th cp my bin p ngun.

: gc iu khin

Biu thc dng ti:Id = d d

d

U E

R

, vi Ed l sut in ng phn ng ng c.

- u im:


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+ Khi dn ch c mt van dn nn st p trong mch van nh nn thch hp vi

phm vi in p lm vic thp.

+ S dng ngun ba pha nn cho php nng cng sut ti ln nhiu ln.

- Nhc im:

+ Cn c bin p ngun c im trung tnh a ra ti, cng sut my bin p

ny ln hn cng sut mt chiu 1,35 ln.

3.2. Chnh lu cu mt pha

Hnh 1.5. S v th lm vic ca chnh lu cu mt pha

Gi tr Ud0 = 0,9U2.

Vi ti l ng c, in cm Ld ln coi dng id l lin tc v phng, c

quy lut iu chnh:

Ud = Ud0cos

Biu thc dng ti:

Id = d dd

U E

R

- u im:

+ C th mc trc tip mch chnh lu vo li in.


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- Nhc im:

+ C hai van dn cng lc nn st p trong mch tng gp i so vi s hnh

tia nn khng thch hp vi ti cn dng ln m in p ra nh.

+ Xung iu khin phi a ti 2 van dn trong cng mt thi im.

3.3. Chnh lu cu ba pha

Hnh 1.6. S v th lm vic ca chnh lu cu ba pha

Gi tr Ud0 = 2,34U2. Vi ti l ng c, in cm Ld ln coi dng id l lin tc v phng, c quy

lut iu chnh:

Ud = Ud0cos

Biu thc dng ti:


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Id = d dd

U E

R

- u im:

+ Cho php u thng vo li in ba pha.

+ p mch rt nh (5,7%).

+ Cng sut bin p xp x cng sut ti, li in t b mo.

+ Phm vi iu chnh cng sut ln.

- Nhc im:

+ C hai van dn cng lc nn st p trong mch tng gp i so vi s hnhtia nn khng thch hp vi cp in p ra ti thp.

T nhng phn tch khi qut trn, chn phng n mch lc l chnh lu cu ba

pha thit k h truyn ng cho my bo ging 7210.


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CHNG II. THIT K MCH LC

1. Nguyn l mch lc

1.1. S

1.2. Gin in p, dng in

Hnh 2.1. S v gin in p, dng in ca chnh lu cu ba pha.


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1.3. Nguyn l lm vic v cc biu thc tnh tonNgun xoay chiu 3 pha c dng :

+ UA = Um Sin()+ UB = Um Sin(2/3)+ UC = Um Sin(+2/3)

Mch van c u thnh hai nhm: nhm katt chung (T1, T3, T5) v nhm

ant chung (T2, T4, T6).

Xung iu khin c pht ln lt theo ng th t t T1 n T6 cch nhau 60 0.

thng mch cn c hai van cng dn, trong mi nhm phi c mt van

tham gia, do 2 van c th t cnh nhau phi c pht xung cng lc. V vy, dngxung l dng xung kp: xung th nht c xc nh theo gc iu khin, xung th hai

m bo iu kin thng mch, nn quy lut van dn nh sau:

- Trong khong + + )(6 2

th van T1 v T6 dn.

- Trong khong5

( + + )2 6

th van T1 v T2 dn.

- Trong khong

5 7

+ )( +6 6

th van T2 v T3 dn.

- Trong khong7 3

+ )( +6 2

th van T3 v T4 dn.

- Trong khong3 11

+ )( +2 6

th van T4 v T5 dn.

- Trong khong 21

6+ )

1( +

6

th van T5 v T6 dn.

Quy lut iu chnh:

Gi tr Ud0 = 2,34U2.

Vi ti l ng c, in cm Ld ln coi dng id l lin tc v phng, c quy

lut iu chnh:

Ud = Ud0cos


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Biu thc dng ti:Id = d d

d

U E

R

1.4. S mch lcS dng mch lc in cm L gim p mch dng in ti.

S mch lc hon chnh (hnh 2.2) gm c: Bin p lc, mch van, mch lc

v mch bo v van bn dn.

Hnh 2.2. S mch lc


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2. Thit k bin p lc

2.1. Tnh cng sut my bin p lc

My bin p lc u /Y do in p pha s cp MBA: U1=380 V

in p pha th cp ca MBA c xc nh nh sau:Phng trnh in p chnh lu khi ti nh mc:

0 min. 2.ddm d Ddm v loc baU U cos U U U U

Trong chn cc tham s nh sau:

+ UDdm l in p nh mc ng c, UDdm = 220V.

+ Vi chnh lu cu 3 pha min = 100 150, chn min = 10

0.

+Uv l st p trn 1 van thyristor trong khong (23)V, chn st p trn 1 van

l 2 V, vi chnh lu cu ba pha ti mt thi im c 2 van dn nn tng st p

trn cc van l 4V.

+ Ulc l st p trn cun khng lc, thng Uloc = (5%10%)Um nn chnUlc = 5%Um = 5%.220 = 11 V+ Uba = Ur + Uxl st p trn in tr v in khng my bin p quy v mtchiu, thng thng Uba = (5%10%)Um, nn chn Uba = 5%.UDm =11 V.

V vy:Uddm = 220 + 2.2 + 11 + 11 = 246 V

m: 0 min 2 min. 2,34ddm d U U cos U cos = 246 V

Do in p pha th cp my bin p: 2 0min

246107( )

2,34. os 2,34. os10ddmUU V

c c

Dng in hiu dng th cp ca my bin p:

2 0,816. dI I

Vi I l dng chnh lu chnh l dng ng c v vy I=I=345 ADo : 2 0,816. 0,816.345 282( )dI I A Dng in hiu dng pha s cp ca my bin p:

I = 1k . I = UU . I = 107380 . 282 = 79,6A


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Tng cng sut my bin p:

Sba = kp.Pd = kp.Udm.Id = 1,05.246.345 = 89113,5 VA = 89,1135 kVA

2.2. Thit k bin p

2.2.1. Tnh s b mch t

Tit din tr QFe ca li thp my bin p c tnh t cng thc :

= . Trong :

Sba - cng sut my bin p c tnh bng [W]

kQ - h s ph thuc phng thc lm mt

+ kQ = 45 nu l bin p du+ kQ = 56 nu l bin p kh

thit k my bin p kh nn chn: kQ = 6

m - s tr ca my bin p; nn:

89113,56 146

3.50

FeQ cm2

ng knh tr:

4. FeFe

Qd

Thay s vo c:

4.14613, 6 [ ]Fed cm

Chun ha ng knh tr chn dFe =14 [cm]

2.2.2. Tnh ton dy qun my bin pThng s cc cun dy cn tnh bao gm s vng v kch thc dy. Loi y qun

s dng l dy ng c in tr sut =0,0000172 [.mm2] , thng s c th c tnhnh sau:


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S vng dy mi pha s cp my bin p:

= 4,44..Trong :

U1 in p pha s cp

QFe tit din s b ca tr

BT t cm ( thng chn trong khong 11,8 T ), chn BT = 1TThay s vo c:

1 4

380W 117

4,44.50.146.10 .1 [vng]

S vng dy mi pha th cp my bin p:

22 1

1

107W .W .117 33

380

U

U [vng]

Tit din dy qun v ng knh dy qun:

-Tit din:

= []Trong : I dng in chy qua cun dy

J mt dng in chy trong my bin p J=(22,75)A/mm2,chn J=2,5 A/mm2-ng knh:

= 4 p dng cho cc cun dy pha s cp v th cp nh sau:

-Tit din v ng knh dy qun s cp:

211

1

11

79,631,8 [ ]

2,5

4 4.31,86, 4 [ ]

3,14

dq

dq

IS mm

J

Sd mm


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-Tit din v ng knh dy qun th cp:

222

2

22

282112,8 [ ]

2,5

44.112,8 12 [ ]3,14

dq

dq

IS mm

J

Sd mm

2.2.3. Kch thc mch t- Din tch ca s: = + -Din tch ca s pha s cp:

21 1 1.W 2.116.31, 8 7377, 6 [ ]cs dqQ k S mm

-Din tch ca s pha th cp:

22 2 2.W 2.33.112,8 7444,8 [ ]cs dqQ k S mm

Do , = + =14822,4 mm2Khi c Qcs cn chn kch thc c bn ca ca s mch t gm chiu cao h v

chiu rng c ca mch t, din tch mch t c tnh Qcs = h.c.

Trong :

Chiu cao ca s mch t: h = m.d ; vi:

m: h s t l, chn m=2,3

d: ng knh tr, d = dFe = 14 [cm], nn:

. 2,3.14 32, 2 [ ]h m d cm

Chn h = 32 [cm]

2.2.4. Kt cu dy qunDy qun c b tr theo chiu dc trc, mi dy qun c qun thnh nhiu lp

dy. Mi lp dy c qun lin tc, cc vng dy qun st nhau. Gia cc lp dy c

ba cch in.S vng dy ca mt lp:

W .gii cn

h hk

d

Trong : h chiu cao ca s mch t


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hg khong cch cch in vi gng

dn - ng knh dy qun k c cch in

kc h s p cht, ly kc = 0,95

-i vi cun dy s cp:

+ s vng dy trn mt lp ca cun s cp:

111

1

2 320 2.6,4w .0,95 45 [ ng]

6, 4ch d

k vd

+ s lp cun s cp:

11

11

w 1173

w 45

n [lp]

-i vi cun dy th cp:

+ s vng dy trn mt lp ca cun th cp:

222

2

2 320 2.12w . .0,95 23[ ]

12ch d

k v ng d

+ s lp cun th cp:

22

22

w 33 2w 23

n [lp]

Chiu dy cun s cp v th cp:

+ s cp: Bd1=(d1 +cd).n1

+ th cp: Bd2=(d2 +cd).n2

Trong : cd chiu dy ba cch in, ly cd = 0,1mm

Thay s vo c:

1 (6, 4 0,1).3 19, 5 [ ]

2 (12 0,1).2 24, 2 [ ]

Bd mm

Bd mm

ng knh trong ca ng cch in:


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Chn ng dy lm bng vt liu cch in c b dy S01 = 0,1 [cm]

Khong cch t tr ti cun dy s cp chn cd01 = 1 [cm]

Khi ng knh trong ca ng cch in l:

Dt = dFe + 2a01 - 2.S01

Thay s vo thu c : Dt = 14 + 2.1 - 2.0,1 = 15,8 [cm]

ng knh trong ca cun s cp:

Dt1 = Dt + 2.S01 =16 [cm]

ng knh ngoi ca cun s cp:

Dn1 = Dt1 + 2.Bd1 = 16+ 2.1,95 = 19,9 [cm]

ng knh trung bnh cun s cp:

1 11

16 19,917, 95 [ ]

2 2t n

tb

D DD cm

ng knh trong cun th cp:

Chn b dy lp cch in gia cun s cp v th cp cd12 = 1[cm]

Nn ng knh trong cun th cp l :

Dt2 = Dn1 + 2.cd12 = 19,9 + 2.1 = 21,9 [cm]

ng knh ngoi cun th cp:Dn2 = Dt2 + 2.Bd2 = 21,9 + 2.2,42 = 26,74 [cm]

ng knh trung bnh cun th cp:

2 22

21,9 26,7424,32[ ]

2 2t n

tb

D DD cm

Chiu di dy qun s cp:

L1 =

.W1.Dtb1 = 3,14.117.17,95 = 6594,5 [cm]

Chiu di dy qun th cp:

L2 = .W2.Dtb2=3,14.33.24,32=2520[cm]2.2.5. in tr my bin p

Chn dy qun l dy ng c in tr sut =0,02133 [.mm2/m]


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23

-in tr cun s cp:

11

1

65,945. 0,02133. 0,0442 [ ]

31,8cu

lR

S

-in tr cun th cp:

22

2

25,2. 0,02133. 0,0048 [ ]

112,8dq

lR

S

-in tr my bin p:

2 2

22 1

1

33. 0,0048 .0, 0442 0, 0083 [ ]

117baW

R R RW

2.2.6. in khng my bin pin khng my bin p c xc nh theo cng thc:

2 71 22

28 .W . . . .103

bkba

R Bd BdX cd

h

Trong :

Rbk bn knh trong ca cun dy th cp

h chiu cao ca s li thp

cd b dy cc cch in cc cun dy vi nhau

tn s gc ca in p

ng knh trung bnh ca cc cun dy:

1 212

16 26,7421, 37 [ ]

2 2t nD DD cm

Nn Rbk= D12 = 21,37[cm]; thay s vo ta c:

2 2 721,37 1,95 2, 4228 .33 . . 0, 01 .10 314.10 0,031[ ]32 3

0,0310,1[ ]

314

ba

baba

X

XL mH

St p trn in khng my bin p:

3. .X ba dU X I

Thay s vo c :3

.0, 032.345 10,5 [V]XU


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24

2.2.7. St p trn my bin p- St p trn in tr my bin p:

.ba ba d Ur R I

Thay s vo c :Urba = 0,0083.345=2,8635 [V]

- St p trn in khng my bin p:

3. .X ba dU X I

Thay s vo c :3

.0, 032.345 10,5 [V]XU

Do , st p trn my bin p:

2 2ba r xU U U

Thay s vo thu c:

2 22,8635 10,5 10,9[ ]baU V

Vy st p trn my bin p thit k trn hon ton ph hp vi st p gi thit ban

u.

3. Chn van thyristor

Khi chn van cn quan tm ti hai ch tiu chnh:

- Ch tiu v dng in: y thng phi tnh c tr s dng in tring bnh lnnht chy qua van, mt khc cng cn quan tm n dng dng in, gi tr dng

in nh, dng qu ti tu theo tng trng hp c th.

- Ch tiu v in p: ch yu l in p ngc t ln van trong qu trnh lm vic.

3.1. Chn van theo ch tiu dng inDng trung bnh qua van c chn theo:

. 3d

V Iv tbv Iv

I

I k I k

Ivk : h s d tr v dng in cho van, chn 2Ivk

tbvI : dng in trung bnh qua van

Id : dng in phn ng nh mc, Id = 345 A


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25

Thay s vo c dng trung bnh qua van:

3452. 230

3VI A

3.2. Chn van theo ch tiu in pin p ca van c chn tho mn iu kin:

Uv > kUv.Ungmax

Trong :

KUv h s d tr v in p cho van, thng ly t 1,72,2 v chn KUv=2Ungmax in p ngc ln nht t ln van, Ungmax= 26U

thay s vo ta c UV >2. 6. 107 = 524Vy nn theo [TL.1] chn van T14-250 do Nga ch to chu c cc iu kin

dng v p trong mch. Cc thng s ca van c cho trong bng sau:

Itb (A) 250

Ung (V) 600

U (V) 1,75

tph (s) 200

Ig (mA) 200

Ug (V) 3,5

di/dt (A/s) 200

du/dt(V/s)

600

Ch thch:

Itb dng in trung bnh cho php.

Ung in p ngc cc i cho php t ln van.

U St p thun trn van.

tph thi gian phc hi tnh cht kho cho van.

Ug in p iu khin.


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26

Ig dng in iu khin.

di/dt tc tng dng cc i cho php ca van.

du/dt tc tng p thun trn van.

4. Tnh chn cun khng san bng4.1. Tnh gi tr cun khng lc

Gi tr in cm xc nh theo cng thc :

L =dm 1

dR

m 2 1sbk

dm 1

dR

m ksb (H) nu ksb 5 (1)

Trong :

Rd l tng tt c cc in tr ti:

Rd = dd

U

I

Vi Ud = Uddm = 246 V

Id = 345 A

nn Rd =246

0,713( )345

d

d

U

I

mm l s ln p mch ca in p chnh lu trong mt chu k in p

ngun xoay chiu, vi chnh lu cu 3 pha th mm =6

1 l tn s gc ca in p xoay chiu, 1 = 2f = 2.3,14.50 = 314 (rad/s)

ksb l h s san bng nh gi hiu qu ca b lc c tnh nh sau:

Ksb = dmvdmr

k

k

Vi kmv l h s p mch u vo

kmr l h s p mch u ra

c hai h s p mch ny u tnh theo biu thc nh ngha:

dmk =1

0

mU

U


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27

H s p mch u vo l ca mch van chnh lu, b thay i theo gc iu

khin . Gc iu khin cng ln th in p ra cng xu v h s p mch cng

tng. Khi tnh ton thng dng th quan h ca h s p mch tng i :

*dmk ( ) = 10( )mUU

(2)

Vi U1m l bin sng hi bc 1 ca in p chnh lu gc iu khin ,

t y suy ra h s p mch thc khi iu chnh vi ny l:

dmk ( ) =1 ( )m

d

U

U

= 1

0

( )

osm

d

U

U c

= *dmk

1

osc (3)

Quan h *dmk = f( ) khi dng in ti lin tc c dng:

2 2 2 2*2.

2 22cos sin cok ( ) s si1 ndm dm mi d dmdmn mm mmk

(4)

Khi gc m nh nht = min th in p trn ti l ln nht :

Udmax = Ud0.cos min = Udm v tng ng vi tc ng c l nh mc:

max = m =mn

9,55 =

1500

9,55= 157 (rad/s)

Khi gc m ln nht = max th in p trn ti l nh nht:

Udmin = Ud0.cosmax v tng ng tc ng c s nh nht min

T D =max

min

=

20

1suy ra min =

max

20=

157

20= 7,85 (rad/s)

(vi D = 20:1 l di iu chnh tc m yu cu cng ngh t ra)

Vy Udmin c xc nh nh sau:

dmin d0 max Dmin v loc baU =U .cos =U +2.U +U +U

Vi UDmin l in p t ln ng c tc nh nht ng vi ti l nh mc.

Dmin u u udm dm min u udmU =E +R I =K +R I nn

dmin d0 max dm min u udm v loc baU =U .cos =K +R I +2.U +U +U

dmin dminmax

d0 2

U U = arcos = arcos

U 2,34.U


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28

Vi dm dm udmdm

U -I .R 220-345.0,05K = = =1,29

157

min = 7,85 (rad/s)

Ru = 0,05 ()Iudm = 345 (A)

vU = 2 (V)

locU = 11 (V)

baU = 11 (V)

Thay s vo phng trnh trn c:

dmin

dminma x

d0

0ma x

U =1,29.7,85 +0,05.345+2.2+11+11 = 53,4 (V)

U 53,4cos = = =0,2136U 250

= 77,7

Khi gc iu khin l max th h s p mch tng ng s ln nht, thay gc iu

khin 0max =77,7 tnh c trn vo c:

2 2 2 2 2 22 2

*

* 2 2cos sin (0, 2136) 6 1 (0, 2136) 0,3351 6 1

1(3) : ( os 0, 2136) 0,335 / 0, 213

k (

6 1,57os

)max md ax dm maxdm

dm d

m

m

mm

Theo k c k c

Nh vy khi cha lc , bin sng hi s ln gp 1,57 ln gi tr trung bnh,

do yu cu cng ngh cn m bo h s p mch nh, chn kmr = 0,1. Do

phi a thm in cm lc L vi h s san bng ca b lc:

Ksb = dmvdmr

k

k=, , = 15,7

T xc nh c in khng lc l:

L =dm 1

dRm

ksb =0,7136.314

.15,7 = 0,00594 H

Chn L = 6 mH, nh vy vi gi tr in cm ny gc iu khin ti a khi

Ud=53,4V, Id = 345/20 = 17,25A s m bo c dng ti ch nhp nh vi bin

khng qu 0,1.Id = 0,1.17,25 = 1,725 A


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29

in tr cun khng lc c tnh theo cng thc:

RCK= r20=

[,.()]Trong :

r20: in tr dy qun nhit 200C

Tmt: nhit mi trng, chn Tmt=500C

T: bin thin nhit , chn T=700C

Thay s vo c:

RCK =/

[,.( )] =

0,022

4.2. Thit k cun khng lc

Kch thc li thpKch thc c s :

2 2442, 6 2, 6 0, 006.354 13, 44[cm]da LI

. Chn a = 14 [cm]

Cc kch thc b,c,h xc nh theo

quan h vi a. Thng thng thng chn

khch thc b,c,h theo a nh sau:

b = (11,5).a

c = (0,60,8).a

h = (23).a

Chn

1, 2 16,8[cm]

0,8 11, 2[ ]

3 42 [ ]

b a

c a cm

h a cm

Hnh 2.4. Kch thc hnh hc li thp

ch E

Tit din li thp : Sth = a.b = 14.16,8 = 235,2 [cm2]

Din tch ca s : Scs = h.c = 42. 11,2 = 470,4 [cm2]

di trung bnh ng sc :


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30

Lth = 2.(a + h + c) = 2.(14 + 42 + 11,2) =134,4[cm]

di trung bnh dy qun :

Ldq = 2.(a + b) +.c = 2(14 + 16,8) + .11,2 = 96,77 [cm]Th tch li thp :

Vth = 2ab(a + h + c) = 2.14.16,8(14 + 42 + 11,2) = 31611 [cm2]

S vng dy ca cun cm :

20 0,023.470,4w 414 414 138, 4 [ ]96,77

cs

dq

r S

Lvng chn w = 138 [vng]

Tnh mt t trng :

100wI 100.138.34535424 [ / ]

134,4

d

th

H A m

L

Tnh cng t cm :

Chnh lu cu ba pha: fm = 6.50 = 300 Hz. Nn :

4 43.10 11.10 2,54.10 [ ]

4,44w. . 4,44.138.300.235,2dm th

UB T

f S

Tnh h s theo H v B :V B < 0,005 [T] nn :

0,83 0,836 6 535424717 .10 717. .10 3, 712.10 [H/m]

1000 1000

H

Tr s in cm nhn c:

2 5 2.w . 3, 712.10 .138 .235, 20,012[ ]

100. 100.134, 4th

ttth

SL H

L

Tit din dy qun:

s = 0,072. = 0,072,. ,, = 101,3 [mm2

]

ng knh dy trn: d = 1,13 = 1,13101,3 = 11,4 [mm] Xc nh khe h ti u:

lkk= 1,6.10-3.w.Id = 1,6.10

-3.138.345 = 76,2 [mm]


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31

V trn ng i mch t c hai on khe h nn ming m s c dy bng

na tr s trn.

lm = 0,5 lkk= 38,1 [mm]

Kch thc cun dy:Dy qun c tit din s = 101,3 [mm2] chn dy qun hnh ch nht c kch

thc 8 x 12,3 =98,4 [mm2].

Chn li cun dy c dy c dy 5 [mm], nn cao s dng qun dy

s l:

hsd = h-2c = 42-1 = 41 cmS vng dy trong mt lp :

w= = = 41Tnh s lp dy:

n == = 3,4 phi qun 4 lpNu ly khong cch gia hai lp dy qun(lp cch in) c l 1[mm] th

dy ca c cun dy s l:

cd = n(d+

c) = 4(0,51+0,1) = 2,44 [cm]

Kim tra kch thc ca s

cd < ccun dy lt trong ca s. Kim tra chnh lch nhit .

Tn tht trong dy qun ng:

Pcu =, ,.() =

, .. ,.( ) = 3432 [W]Tng din tch b mt cun dy:

Scu= 2hsd(a+b+. cd) + 1,4. cd(.0cd + 2.a)Scu= 2.41(14+16,8+.2,44)+1,4.2,44(.2,44 + 2.14)=3275,7 [cm2]

H s pht nhit :

= 1,03.10-3 = 1,03.10-3 = 7,25.10-4 [w/0C.cm2]


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chnh lch nhit :

T = = ,., = 14450c chnh lch nhit ln hn mc cho php nn ta c th chn lm mt cng

bc.

5. Tnh ton bo v mch lc

5.1. Bo v qu dngC hai kiu bo v qu dng l bo v qu dng ngn hn v bo v qu dng lu

di. i vi cc mch c Thyristor cn c mch bo v tc tng dng in qua van.

5.1.1. Bo v dng qu ti lu di

Khi u song song nhiu van cng loi tng dng tng, do c tnh VA cavan khc nhau dn n s phn dng cho mi van khng u nn c van s chu dng

ln hn tnh ton. chng hin tng ny phi dng cc phn t h tr phn b dng

cho u bng in tr hoc in cm.

Hnh 2.5. Cc phng php chia dng khi van u song song

5.1.2. Bo v qu dng in dng ngn hnKhi mch c s c, dng in trong mch chnh lu tng nhanh v thng ko

di c 10ms mc d cc phn t bo v tc ng. Thng trong thng s ca vanbao gi cng c gi tr dng in m van c kh nng chu c trong 10ms, gi tr

ny ln hn tr s trung bnh cho php t 8 ti 10 ln. V vy nu cc van c chn

c tr s dng in ny nh hn dng s c qua van trong thc t th hoc phi thay


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loi van khc hoc phi a thm phn t hn ch s tng trng dng s c xung

n mc cho php ca van. Cc phng php bo v qu dng in ngn hn sau y:

- Bo v bng cu ch.

- Bo v bng aptmat.- Bo v bng cch ngt xung iu khin.

- Bo v bng cch chuyn mch chnh lu sang ch nghch lu ph thuc.

- Bo v bng mch kho cng bc cc Thyristor lc:

Hnh 2.6. Bo v bng kho cng bc van

Bnh thng t C c np n tr s Uc. Khi c s c, mch bo v pht xung

m T1, t C s t in p ngc ln cc van mch lc lm chng kho li, ngt dng

s c. Tr s t in chn theo biu thc:.sc ph

c

I tC

U ,

Trong :

Isc: dng s c;

Tph: thi gian phc hi tnh cht kho cho van

Uc: in p t c np t ngun ngoi.

5.1.3. Bo v tc tng dng di/dt cho ThyristorDo b chnh lu thit k c bin p lc nn vic bo v tc tng dng cho

Thyristor c m bo.


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34

5.2. Bo v qu in p

5.2.1. Cc nguyn nhn gy qu p

- Qu p t li in a ti.

- Qu p do ng ngt cc khi chc nng ca bn thn b chnh lu.- Qu p do hin tng chuyn mch gia cc van khi lm vic.

5.2.2. Cc bin php bo v qu p

- Bo v qu p do pha ngun xoay chiu gy ra: dng mch RC (hnh 2.4a)

- Bo v cc xung p trn van: dng mch RC mc song song van (hnh 2.4b).

a)b)

Hnh 2.7. Cc mch bo v qu p cho chnh lu

Tnh bo v van bn dn dng mch RC mc song song vi van.in p ngc ln nht khi hot ng: Ungmax=Udym=2,45.U2=2,45.107=262,15VVy h s qu p: k = Ungcp/Ungmax = 600/262,15 = 2,29.

Hnh 2.8. th xc nh RC mc song song Thyristor

Vi h s k = 2,29 tra th hnh 2.6 c:

C*min = 0,28 ; R*

max = 2,1 ; R*

min = 1


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35

C La = Lba =0,1 mH. Theo s thay th giai on chuyn mch trng dn

gia 2 van ta c tc tng dng ln nht s tng ng in p bng bin in p

dy Udym nn:

y 23

ax2,45. 2,45.107 1,311 /2 2 2.0,1.10

d m

m a a

U Udi A sdt L L

Vi chnh lu cu ba pha theo [TL.1] c quan h quy i:

L= 2La ; R= 0,6R ; C =1,67C

Hnh 2.9. th Q tch lu trong van

Vi dng Id = 345 A v di/dt = 1,311A/s tra th hnh 2.5 c in tch tn ti

trong van trc khi kho l: Q = 120 As. T y xc nh c cc phn t bo v: T C:

' *min min

max

2 2.120. .0, 28 0, 256

262,15ng

QC C F

U nn C = C/1,67 = 0,15 F

chn C = 0,22 F

in tr R:' 3

max*min 6

' 3max*

max 6

. 2.0,1.10 .262,15' 1. 14,8

2 2.120.10. 2.0,1.10 .262,15

' 2,1. 312 2.120.10

a ng

a ng

L UR R

QL U

R RQ

suy ra:14,8 R=0,6R 31 nn 24,7 R 51,7

Chn R = 33.


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36

CHNG III. THIT K MCH TO XUNG IU KHIN

cho cc van ca b bin i m ti thi im mong mun ta cn c mch pht

xung iu khin a n m cc Thyristor ti cc thi im yu cu nh: bin , tn s,cng sut v thi gian tn ti m chc chn cc van vi mi ti m s gp phi khi

lm vic nn s dng phng php iu khin dc:

Hnh 3.1. Nguyn tc iu khin dc

Trong :

- Khu ng b (B) c hai chc nng:

+ m bo quan h v gc pha c nh vi in p van lc nhm xc nh im gc

tnh gc iu khin

+ Hnh thnh in p c dng ph hp lm xung nhp cho hot ng ca khu to

in p ta pha sau n.

- Uta l khu to in p ta dng c nh, thng c dng rng ca.

- Khu so snh (SS) xc nh im cn bng ca hai in p U ta v Uk pht ng

khu to xung.

- Khu to dng xung (DX) nhm to ra xung c dng ph hp m chc tiristor. n

s dng khu to xung kp.

- Khu khuch i xung (KX) c nhim v khuch i cng sut xung sau khu to

dng xung mnh m van lc.


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37

1. Khu ng b

1.1. S (hnh 3.2a)

1.2. Gin xung (hnh 3.2b)

a)b)

Hnh 3.2. To xung nhp ng b kt hp chnh lu v OA

1.3. Gii thiu s y s dng mch chnh lu hai na chu k s dng mch chnh lu hnh tia 2 pha

dng 2 it D1, D2 v ti cho chnh lu ny l Ro. in p chnh lu c a ti ca

(+) ca khuch i thut ton OA1 so snh vi in p ngng Ung ly t b phn p

iu chnh gia bin tr P1v in tr R4.

1.4. Nguyn l lm vic

in p ng b tun theo quan h sau:

Ub = A0(U+ - U-) = A0(Ucl Ung )

Do : Nu Ucl > Ung th Ub dng v bng in p bo ha ca OA.

tng t nu Ucl < Ung th Ub = -Ubh.

-T 0 n 1 th Ucl < Ung nn in p Ub m v bng in p m bo ha caOA1


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38

-T 1 n 2 th Ucl > Ung nn in p Ub dng v bng in p dng bo

ha ca OA1

-T 2 n 3 th Ucl < Ung nn in p Ub m v bng in p m bo ha ca

OA1-T 3 n 4 th Ucl > Ung nn in p Ub dng v bng in p dng bo

ha ca OA1

T thu c th in p Ub c dng xung vung nh trn hnh v. im

giao nhau ca Ucl v Ung l im chuyn trng thi ca in p Ub, cc im giao nhau

ny chnh l hai im xc nh gii hn ca gc iu khin ( vi=1 v = 2).2. Khu to in p ta

2.1. S (hnh 3.3a)


a)

b)

Hnh 3.3. To rng ca i xung hai na chu k bng OA

2.3. Gii thiu s y l s to rng ca tuyn tnh i xung trong c hai na chu k ca in p

li xoay chiu thc hin theo nguyn tc dng chnh lu it bin in p xoay chiu

thnh in p ch c mt cc tnh. Cc tnh ny phi thch hp cho giai on to rng


39/62

39

ca, ng thi mch thm vo Dz m bo phc hi trng thi ban u cho t in trong

thi gian ngn khi in p li gn im qua khng.

in p ng b ch c hai tr s m v dng bng gi tr bo ho Ubh. TngOA2 lm nhim v to rng ca. Bng cch iu chnh P1 ta thay i ngng ny lmthay i c quan h gia thi gian to rng ca trc (tng ng vi thi gian phng ca

t tp) v thi gian hi phc tn theo yu cu.

2.4. Nguyn l lm vic

Mch to xung rng ca dng khuch i thut ton (hnh 3.3) c xy dng da

trn nguyn tc s dng mch tch phn. Qu trnh phng np ca t thc hin bng

ngun hai cc tnh. to ra c xung rng ca tuyn tnh th dng qua t phi dm

bo khng i theo thi gian

Qu trnh np t :

max 1 201 1

1 1.rc

U

n c nU i dt I I t C C

1 25 6

;1

bhU EI IR P R

suy ra: max 1 5 6

1

1bh

rc

U E

U C R P R

qu trnh np nhanh thit k R6 >> R5 dn ti I1 >> I2 dng I2 rt nh,

n gin b qua dng I2 coi dng np cho t ch l I1.

Qu trnh phng ca t :

max

0

21

1.

rcp pU

U I tC

suy ra: max 21

1

.rc pU I tC

-T 0 n 1 th Ub = -Ubh nn t C1 c np n in p Urcmax

-T 1 n 2 th Ub = Ubh iot D9 kho, t C1 phng in v Urc = 0

-T 2 n 3 th Ub = -Ubh nn t C1 c np n in p Urcmax

-T 3 n 4 th Ub = Ubh iot D9 kho, t C1 phng in v Urc = 0


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Kt qu thu c th in p rng ca i xung hai na chu k nh hnh 3.3.

3. Khu so snh

3.1. S (hnh 3.4a)


a)b)

Hnh 3.4. So snh hai ca dng khuch i thut ton OA

3.3. Gii thiu s

S dng mch so snh kiu hai ca dng khuch i thut ton OA. Trong mch so

snh kiu ny, hai in p cn so snh l Urc c to ra t khu to in p ta v Uk

c a ti hai cc khc nhau ca OA. Ty thuc in p ta v in p iu khin

c a vo ca no m in p ra Uss xut hin sn xung m hay dng thi im

cn bng gi tr gia chng.

3.4. Nguyn l lm vic

in p ra s tun theo quy lut:

Ura=K0(u+ - u-), K0 l h s khuch i ca OA

V in p iu khin a vo ca (+) cn in p ta a vo ca (-): u+ uk v

u- uta (urc) th in p ra l:

Ura = K0(u+ - u-) = K0(uk- uta)

Do khi uk> uta th ura = + Ubh v ngc li uk< uta th ura = - Ubh

in p so snh Uss thu c c dng xung vung nh hnh 3.4.


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41

4. Khu trn xung v khuch i xung (KX)

4.1. Khu trn xung

y mch trn xung thc hin chc nng ca mch logic AND 3 cng vo ly t

+ Tn hiu ra t khu so snh: s cho php hay cm khu khuch i xung nhn

xung tn s cao pht t b to dao ng ti n.

+ Tn hiu ra t b to dao ng tn s cao.

+ Tn hiu ra ca khu tch xung dng OA.

4.1.1. Tn hiu ra t khu so snh

a) b)

Hnh 3.5. Mch to xung dng t u ra ca khu so snh.

Do cc mch IC logic ch lm vic vi ngun mt cc tnh nn khi ghp IC logic

vi OA (nu OA dng ngun nui hai cc tnh) phi chn cc xung m a t OA n

ca vo ca IC nh mch ghp RD (in tr - it).Tn hiu xung nhn c th

hnh 3.4b.

4.1.2. Mch to xung tn s cao


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Hnh 3.6. To dao ng dng OA

+ Gi s thi im 0, in p ra ca khuch i thut ton t gi tr cc i v c

gi tr Udd +E.+ Thng qua mch phn hi R13, R14 th u vo ca (+) ca khuch i thut ton

s c tn hiu phn hi 0 1817 18

.E

U RR R

duy tr cho khuch i thut ton nm vng

bo ha dng. Lc ny t C2 c np thng qua R11

ti gi tr E .

+ Khi t = t1, in p Uc t gi tr U0, khuch i thut ton lt trng thi v c gi

tr Udd - E+ in p Uc khng th thay i t ngt v lc ny t C2 phng in qua R11

+ thi im t = t2, khi Uc = 0 1817 18

.E

U RR R

, khuch i thut ton li lt trng

thi v Udd = + E. Qu trnh c tip tc lp li.

Chu k dao ng

T = 2R19Cln(1+ 2R18/R17).

Chn tn s dao ng 10kHz nn chu k to xung T =3

1 1100

10.10s

f


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4.1.3. Mch tch xung

Hnh 3.7. Mch tch xung dng OA

y mch tch xung dng OA phn bit chnh xc hai na chu k in p li

khi n qua im khng, tch hai knh l hai tn hiu m 2 van trong mt pha. Chn cc

xung m a t OA n ca vo ca IC nh mch ghp RD (in tr R7 it D7).

- Nguyn l hot ng:

+ in p ng pha ca (+) c so snh vi 0 ca (-).+ Trong khong t (0 - ): > 0 hay > , in p ra sau OA l + Ubh nn

it D7 kho nn c gi tr l + .+ Trong khong t ( - 2): < 0 hay < , in p sau OA l Ubh nn

it D7 dn nn Utx c gi tr l 0

Vy in p ra sau mch tch xung c dng xung ch nht mt cc tnh rngxung 180 trng vi na chu k dng ca mch ng pha s l tn hin chn knh chovan u catt chung.


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4.3. S ca khu trn xung v khuch i xung

Hnh 3.9. S khu trn xung v khuch i xung

Khu KX c nhim v tng cng sut do khu to dng xung hnh thnh n mc

mnh m van lc. y s dng khuch i xung ghp bng bin p c u im

d dng cch li mch iu khin v mch lc. m bo h s khuch i dng cn

thit tng khuch i u theo kiu Dalintn.

Trong hnh 3.9, Rbax

lm nhim v tiu tn nng lng cun dy trong giai on

kho cc bng bn dn. Khi T2 kho dng in qua bin p xung s chy vng qua 11

Rbax nn nng lng s c tiu tn trn in tr ny. Gi tr Rbax thng c chn t

kh nng dn dng ti a cho php ca T2:

Rbax >

Tuy nhin Rbax mc ni tip vi cun s cp BAX nn khi dn Rbax s lm gim p

t vo BAX, gi in p ban u trn BAX bng ngun ECS c th a thm t Cbax

vo, lc trong giai on T2 kho t in phi kp np n tr s bng ngun, iu kin

tnh tr s t in:

Cbax < vi:

tn : thi gian ngh gia hai xung lin nhau ca xung chm


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Hnh 3.10. Mt knh ca mch iu khin hai na chu k c khu tch xung cho hai

thyristor cng mt pha mch lc.

5. Tnh ton thng s mch iu khin phn trn chn van T14-250 do Nga ch to vi cc thng s ca van c cho

trong bng sau:

Itb Ung U Tph Ig Ug dt

di

dt

du

250 A 600 V 1,75 200 200mA 3,5 200 600

Nh vy mch pht xung phi m bo tha mn hai thng s:

Ig = 200 mA = 0,2A

Ug = 3,5V

m van chc chn cn thit k mch iu khin c U g = 5V, Ig = 300 mA


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5.1. Tnh ton khu khuch i xung

* Tnh ton bin p xung :

Cc bin p xung c h s bin p = 2. in p, dng in th cp (in p,dng in iu khin Ug=5V, Ig=300mA), ly tn s pht xung f = 10 kHz.Ta chn li

thp ferit dng ch E. Bin p hot ng vng t ha mt phn nn chn bin thin

cng t trng B = 0,2 T; bin thin mt t cm H = 30 A/m.

Kch thc bin p xung c tnh theo cng thc:

V =.....

Vi B : bin thin cng t trng (T) .

H : bin thin mt t cm (H/m) .: rng mt xung (s).: st p xung cho php, thng ly t 0,1 n 0,2.

kba, U2, I2 tng ng l h s bin p xung, tr s hiu dng in p v dng

in th cp bin p xung.

Chn tn s pht xung fx =10 kHz. Do , chu k xung l:

63

1 1100.10 ( )10.10x xT sf

rng ca xung l: tx = 6 61 1

.100.10 50.10 ( )2 2x

T s

Nn kch thc ca bin p xung:

V =..... = ..,. ..,,. = 1,25.10-6 m3 = 1,25 cm3

Tra bng cho trng hp t ho mt phn chn li hnh tr k hiu 813E343 c tit

din li tng ng bng 0,412 cm2. Vy s vng dy cun s cp:

w1 =.. = . .

,., . = 60 vng, suy ra w2 = 60/2 = 30 vng.

Bin p xung c t s k=2, tham s in p v dng in cun s cp l:


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U1 = k.Ug= 2.5=10 V

I1 = Ig/k= 0,3/2 =0,15 A

Ngun cng sut phi c tr s ln hn U1 b st p trn in tr v vy chn

ECS =15V. T gi tr ECS v I1 chn bng T2 v T4 loi BD135 c tham s Uce = 45V,

Icmax = 1,5A, tra bng c min=40. Chn bng T1 v T3 l loi BC107 c Uce = 45V;Icmax = 0,1A,min=110. C:

Rbax > = , = 10 , chn Rbax = 12

Kim tra st p trn in tr ny khi bng dn dng:

URbax = I1.Rbax = 0,15.12 = 1,8 V ;

nn in p cn trn BAX l:

U1 = ECS URbax = 15 1,8 = 13,2 V v ln hn 10V nn t yu cu.

Tnh chn in tr R13 v R14:

in tr trn c chn theo cng thc:

R13=R14 = ..,., = 36,67 kVy chn in tr R13 =R14 = 33 k

in tr R15, R16 lm nhim v cch li cc bazo ca transistor T2, T4 vi t,

chn l 10k.

Cc it D11, D12, D13, D14, D15, D16 bo v bin p xung chn loi 1N4002.

5.2. Tnh ton khu trn xung

Chn IC logic cng AND l loi 4073(1 IC c 3 cng AND 3 u vo) c ngun

nui l 5V (Ura = 5V tng ng vi mc logic 1).

5.2.1. Tnh chn khu tch xung


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+ Cc OA khu ny chn loi TL081C

+ Cc diot D3,D4,D5,D6 lm nhim v bo v OA chn loi 1N4002.

+ in tr R2, R3 vo OPAM, vi = 100, chn R2 = R3 = 680.+ Cc cp (R7+D7), (R8+ D8) loi b phn m ca xung u ra OPAM. Chn D7,

D8 l loi 1N4002; in tr R7= R8= 15k.

5.2.2. Tnh chn khu to dao ng

Chn t C = 10nF ; m T= 2R19C.ln(1+2R18/R17), chn in tr R17 = 2R18 suy

ra T=1,4.RC nn: in tr R19 =, = .

,.. = 7,1.103

Vy chn in tr R19 =8,2k.

5.3. Tnh ton khu so snh

Chn in tr R9 = R10 = 10 k

(in tr R12 + it D10 ) chn cc xung m n ca vo IC nn chn in tr

R12 = 15 k, it D10 vn chn loi 1N4002.

5.4. Tnh ton khu to in p ta

Chn E = 12V, Up =11 V, f=50 Hz.

Chn OA loi TL082 cha hai OA trong mt v IC Thi gian t C phng chnh l khong thi gian tng ng vi phm vi iu

chnh gc iu khin , nh bit min = 100 c ngha l phm vi iu chnhgc iu khin l 1700 quy i sang thi gian l:

tp =. = 9,44 ms

Chn it n p BZX79 c UDz = 10V;

Chn t C = 220nF;

Tnh in tr R6:

R6 =.. =

., ..,. = 51,5 kChn mt in tr 39 k ni tip vi bin tr 20 k vo v tr R6


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- Tnh in tr R5 theo cng thc:

R5,. Thi gian t C np in l: tn = T/2-tp = 10-9,44 = 0,56 ms,in p bo ho ca OA l: Ubh = E - 1,5 = 12 1,5 = 10,5 V. Vy:

R5 ,. =,,,. . , . ,. = 2,5 k; chn in tr R6 = 2 k

6.5. Tnh ton khu to xung ng b

+ Nhm chnh lu tia hai pha vi hai it D1,D2 c in p vo l in p

ng pha vi tr s hiu dng 11 V, nn in p ngc ln nht t ln van l:Ung.max = 22.Up = 2.1,414.11 = 31,11 V

Chn it D1,D2 l loi 1N4002 vi tham s: Itb = 1A, Ung.max =100V, in

tr ti cho chnh lu R0 = 1k;

+ Chn it D9 loi 1N4002;

+ Mch so snh to xung ng b. Chn in tr R1 = 15 k

Gc iu khin nh nht min = 100 th in p ngng s bng:Ung = 2.Upsinmin = 2.11.sin100 = 2,7 V.

Tuy nhin nu tnh st p trn it chnh lu th ngng ny phi gim i

0,7V do Ung c gi tr xp x 2V

Chn dng qua phn p (in tr R4 +bin tr P1) l 1mA, tng tr ca b

phn p l :

R = = . = 12 k

T y chn phn p gm in tr R4 = 10 k v bin tr P1 = 2 k.


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CHNG IV. TNG HP H V M PHNG

Thng s ng c:

+ Cng sut nh mc: = 67,5 kW.+ in p nh mc: = 220 V.+ Dng in nh mc: = 345 A.+ Tc nh mc: = 1500 vng/pht.

Hay = 2../60 = 2. .1500/60 = 157 rad/s.+ in tr phn ng: = 0,05 m.+ M men qun tnh:= 7 Kg.+ S i cc p = 2

1. Tnh ton thng s

1.1. ng c1.1.1. Tnh in tr tng mch phn ng R

+ in tr phn ng ng c (): R = 0,05

+ in tr bin p: RBA = 0,0083

+ in tr cun khng lc: RCK= 0,022

Nn R = R + RBA + RCK= 0,05 + 0,0083 + 0,022 = 0,0803

1.1.2. Tnh in cm tng mch phn ng L

+ in cm phn ng : L = DdmLDdm dm

Uk .I .p.n

Trong kL l h s ly gi tr 5,5 5,7 i vi my khng b, kL ly gi tr

1,4 1,9 i vi my c cun b, p l s i cc; nn:


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L =3Ddm

L

Ddm dm

U 220k . = 5,5. =1,17.10

I .p.n 345.2.1500H =1,17 mH

+ in cm ca cun khng lc: LCK= 6 mH

+ in cm my bin p: LBA = 0,1 mH

Nn L = L + LCK+ LBA = 1,17 +6 +0,1 = 7,27 mH

1.1.3. Tnh hng s thi gian in t T

T =37,27.10

0,0910,0803

u

u

Ls

R

1.1.4. Hng s thi gian c TcC : K.m = 1,29

Hng s thi gian qun tnh ca h : JHT = 2.J = 2.7 = 14 Kgm

2

nn: Tc = 2 2. 0,0803.14

0,675( ) (1,29)

HT

dm

R Js

k

1.2. B bin i

Hm truyn ca b bin i:

G = d1

b

v

K

T p

Vi ,: h s khuch i , hng s thi gian ca b bin i.ly Uk=10V: h s khuch i b bin i = = = 24,6

= 1/2mf , m: s ln p mch ca chnh lu.

Vy hng s thi gian b bin i Tv = 1/(2.6.50)=1,67 ms

2. Tng hp mch vng dng in

H thng truyn ng in ng c quay chi tit my bo ging c hng s thi

gian c hc Tc rt ln so vi hng s thi gian in t ca mch phn ng T nn ta c


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th coi sc in ng ca ng c khng nh hng n qu trnh iu chnh ca mch

vng dng in.

hnh 4.1. S khi mch vng iu chnh dng in

Trong Tk, Tv, T, Ti l cc hng s thi gian ca mch iu khin chnh lu, bbin i, phn ng v sensor dng in. Ri l b iu chnh dng in, BB dng chnh

lu cu ba pha, Si l khu o dng in.

T s trn ta c hm truyn ca i tng mch vng iu chnh dng in :

. /

(1 )(1 )(1 )(1 )bd i u

oidk v i u

K K RS p

pT pT pT pT

Trong hng s thi gian si i v dkT T T T


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53

2 2

2 2

11 2 2

( ). / 1

(1 )(1 )(1 ) 1 2 2

ibd i u

si u

p pR p

K K R

pT pT p p

Chn hng s thi gian = siT . Vy ta c hm truyn ca b iu chnh dng in:

1 . 1( ) (1 )

2 . . . / 2. . .u u u

ibd i si u bd i si u

pT R TR p

p K K T R K K T pT

, ( )iR p l khu tch phn ti l (PI).

Hm truyn h kn ca i tng mch vng dng in:pT

Fsi

Ki 21

1

(b qua

thnh phn bc cao)

-Tnh ton tham s hm truyn ca b iu chnh dng in :

Hm truyn khu o dng in:

Gi =1

i

i

K

T p

Vi: Ki: h s khuch i khu o dng

Ti: hng s thi gian ca khu o dng

Chn hng s thi gian khu o dng T i = Tk= 0,0005 < T nn:

H s khuch i khu o dng:

ax

100,0116

2, 5. 2, 5.345i dk

im dm

U UK

I I

Hng s thi gian Tsi:

si i v dkT T T T =0.0005+0.00167+0.0005 = 2,67.310 s

Vy hm truyn ca b iu khin dng in:

iR 30, 0803.0, 091 1 1 52, 7

(1 ) 4,8(1 ) 4,82.24, 6.0, 0116.2, 67.10 0,091p 0,091p p


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- Thit k cu trc b iu chnh dng in:

Hnh 4.2. Cu trc b iu chnh dng in

C:11

( )2 . . . /

uRi R

R bd i si u

pTpR p K

p p K K T R

;

vi: 13

..2 . 0, 019

. 0.091

bd iRsi

R u

R u

K KT R C

K R

T R C

Chn t C = 2,2F nn: in tr R3 = 41,36.103 = 41,36 k

in tr R1 = 8,64.103 = 8,64 k

in tr R2 tho mn: 1

2

id

i

UR

R U

suy ra 2 1.i

id

UR R

U

V tng hp b iu chnh dng in Ri(p) theo tiu chun ti u mun l v sai

cp mt vi tn hiu t nn Ui = Ui = 10 V suy ra in tr R2= R1=8,64.103 = 8,64 k

3. Tng hp mch vng tc

Hnh 4.3. S mch vng iu chnh tc

Rw HCD k

Mc

Uw

Uw

I M w


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Trong :

Rw: B iu chnh tc

HCD: khu hn ch dng

pTK

1: hm truyn ca khu o tc

Vi K : h s khuch i khu o tc

T : hng s thi gian khu o tc

Sow l hm truyn ca i tng mch vng iu chnh tc :

.1/

(1 )(1 2 )

u io

c si

K R KS

K T p T p T p

Do siTT ; l cc hng s thi gian nh nn ta t TTT sis 2

Nn :.1/

(1 )u i

oc s

K R KS

K T p T p

p dng tiu chun ti u mun c hm truyn ca h thng kn nh sau:

w 2 2

1( )

1 2 2oMF p

p p

Mt khc trn hnh 4.3 c:

ww

w

( ). ( )( )

1 ( ). ( )i o

oMi o

R p S pF p

R p S p

nn:w w

w w w w w

( ) ( )( )

( ) ( ). ( ) ( )[1 ( )]oM oM

io oM o o oM

F p F pR p

S p F p S p S p F p

Chn sT . Vy hm truyn ca b iu chnh tc :

. .( )

. .2i c

u s

K K TR p

K R T

, Rw(p) l b iu khin t l P


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-Tnh ton tham s hm truyn ca b iu chnh tc :H s khuch i khu o tc :

100,0637

157dm

UK

Chn hng s thi gian ca khu o tc : 0,001[ ]T s

Hng s thi gian Ts:

2. 2.0, 00267 0, 001 0, 00634 [ ]s siT T T s

Vy hm truyn ca b iu chnh tc :

. . 0,0116.1,225.0,749( ) 164

. .2 0,0637.0,0803.2.0,00634i c

u s

K K TR p

K R T

- Thit k cu trc b iu chnh tc :

Hnh 4.4. Cu trc b iu chnh tc

KR: l h s khuch i b iu chnh.

T hnh 4.4 c:

3 3 3 1w w w w

1 2 1 2

3w

1

. . .

( )

w w w wid d d

w w w w

wR

w

R R R RU U U U U

R R R R

RK R p

R

m:. . 0,0116.1,225.0,749

( ) 164. .2 0,0637.0,0803.2.0,00634i c

u s

K K TR p

K R T

Chn in tr R1w = 2,7 k suy ra in tr R3w = 2,7.164 = 442.8 k


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in tr R2 tho mn: w12 w

dUR

R U suy ra w2 1

w

.d

UR R

U

V tng hp b iu chnh dng in Ri(p) theo tiu chun ti u mun l v sai

cp mt vi tn hiu t nn Uw = Uw suy ra R2 = R1 = 2,7 k.

Do KR = 164 kh ln nn thng dng 2 tng khuch i. Tng th nht khuch i

10 ln, tng th hai khuch i 16 ln.

4. M phng h thng

gim nh hng ca qu trnh qu ti h thng, thit k khu to tn hiu t

nh sau:

Uwt (t) =

5 [0,2]10 > 2

Hnh 4.5. Tn hiu t

S m phng:


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Khu hn ch dng (HCD) c nhim v hn ch dng khi khi ng khng vt

qu (2,5 3)Im .

Momen nh mc ca ng c:

Mm = Pm/wm = 67,5.103/157 = 430 N.m

Sau khong thi gian 3s t lc khi ng tin hnh ng ti vi M c=Mm = 430 N.m

vo h thng thu c kt qu nh sau:

Hnh 4.6. th dng in

0 2 4 6 8 10-200

0

200

400

600

800

1000


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Hnh 4.7. th tc

Hnh 4.8. Sai lch dng in

0 2 4 6 8 100

20

40

60

80

100

120

140

160

180

200

0 2 4 6 8 10-4

-3

-2

-1

0

1

2

3

4


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Hnh 4.10. Sai lch tc

Nhn thy th p ng tc v dng in n nh, thi gian qu kh nh

(di 1s). Khi ng ti trong qu trnh qu c st tc v qu iu chnh dng

nhng trong gi tr cho php. i vi tc mc d tn ti sai lch nhng l rt nh

(0,016%) nn c th chp nhn c. khc phc sai lch ny th cn tng hp b iu

chnh tc l b PI. Vy b iu khin p ng c yu cu.

0 2 4 6 8 10-0.01

0

0.01

0.02

0.03

0.04

0.05

0.06


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TI LIU THAM KHO

1. Phm Quc Hi. Hng Dn Thit K in T Cng Sut. Nh xut bn Khoa

hc v k thut. H Ni, 2009.2. L Vn Doanh, Nguyn Th cng, Trn Vn Thnh.in t cng sut : l thuyt

thit K - ng dng. Nh xut bn Khoa hc v k thut. H Ni , 2005.

3. V Minh Chnh, Phm Quc Hi, Trn Trng Minh.in t cng sut.Nh xut

bn Khoa hc v k thut. H Ni, 2007.

4. Bi Quc Khnh, Nguyn Vn Lin, Phm Quc Hi, Dng Vn Ngh. iu

chnh t ng truyn ng in. Nh xut bn Khoa hc v k thut. H Ni,

2012.5. Nguyn Mnh Tin, V Quang Hi. Trang b in in t my gia cng kim

loi. Nh xut bn Khoa hc v k thut. H Ni, 2007.


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